To understand the behavior of the function $f(z) = e^{-\left(\frac{2}{z-1}\right)}$ at $z = 1$, we examine its Laurent series expansion around $z = 1$.

Analyzing the Singularity at z = 1

Let $w = \frac{2}{z-1}$. As $z \to 1$, the term $z-1 \to 0$, and thus $w \to \infty$. The function becomes $f(z) = e^{-w}$.

The Taylor series for $e^x$ around $x=0$ is $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.

Substituting $x = -\frac{2}{z-1}$, we get the Laurent series for $f(z)$ around $z=1$: $f(z) = e^{-\left(\frac{2}{z-1}\right)} = \sum_{n=0}^{\infty} \frac{\left(-\frac{2}{z-1}\right)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^n}{n!} (z-1)^{-n}$ Expanding this series: $f(z) = \frac{(-2)^0}{0!} (z-1)^0 + \frac{(-2)^1}{1!} (z-1)^{-1} + \frac{(-2)^2}{2!} (z-1)^{-2} + \dots$ $f(z) = 1 - \frac{2}{z-1} + \frac{2}{(z-1)^2} - \frac{4}{3(z-1)^3} + \dots$

This series contains infinitely many terms with negative powers of $(z-1)$ (i.e., terms like $(z-1)^{-1}, (z-1)^{-2}, \dots$). This structure indicates that the singularity at $z = 1$ is an essential singularity.

Calculating the Residue at z = 1

The residue of $f(z)$ at an isolated singularity $z_0$ is the coefficient of the $(z-z_0)^{-1}$ term in its Laurent series expansion. In this case, $z_0 = 1$.

From the Laurent series derived above:

The term corresponding to $(z-1)^{-1}$ occurs when $n=1$. The coefficient is:

Therefore, the residue of the function $f(z) = e^{-\left(\frac{2}{z-1}\right)}$ at $z = 1$ is $-2$.

Conclusion

• The presence of infinitely many terms with negative powers in the Laurent series confirms an essential singularity at $z=1$.
• The coefficient of the $(z-1)^{-1}$ term is $-2$, confirming the residue is $-2$ at $z=1$.